Potential from continuous charge

Superposition for potential is ordinary addition: the potential at P from several charges is V = k Σ qᵢ/rᵢ, carrying signs but no directions. Let the number of charges grow and their size shrink and that sum becomes V = k ∫ dq/r. The condition for chopping is simple — each piece must be small enough that every part of it lies essentially the same distance r from P — and the pay-off is large. The matching field calculation needs three integrals, one per component, plus a symmetry argument for which of them vanish. Here three contributions of +12 V, −5 V and +3 V make +10 V, and there is nothing else to say. One warning travels with the gain: V is a number, so it cannot carry a direction. Recovering E means differentiating the answer afterwards, as a separate step.

Charge along a wire is described by a linear density λ = dq/dℓ in C m⁻¹, over a surface by σ = dq/dA in C m⁻², and through a volume by ρ = dq/dV in C m⁻³. The element you write is the whole set-up. On a straight rod along x, dq = λ dx and the variable is x. On a disc it pays to use annular rings rather than little squares: the ring between radii s and s + ds has area dA = 2πs ds, so dq = σ 2πs ds and one variable does the work of two. A thin spherical shell of radius r and thickness dr gives dq = ρ 4πr² dr. Uniform densities come straight from the totals — λ = Q/L, σ = Q/πR², ρ = 3Q/4πR³ — while a non-uniform one stays inside the integral: a rod with λ(x) = λ₀x/L contributes dq = (λ₀x/L) dx, and that factor of x must be carried, never averaged away first.

This is where the calculation is won or lost. In dV = k dq/r, the symbol r is the distance from the element to the field point, and it has to be re-expressed in the integration variable. For a rod along x with P on the same line, a distance d beyond the near end, the element at x sits at r = d + x. Move P onto the perpendicular bisector instead, a height y above the rod, and with x now measured from the midpoint r = √(x² + y²). And for a ring of radius a with P on the axis at height z, every element is at r = √(z² + a²) — a constant. A constant pulls straight out: V = k/√(z² + a²) ∫ dq = kQ/√(z² + a²). With a = 5.00 cm, Q = +4.00 nC and z = 12.0 cm, r = 13.0 cm and V = 277 V. Notice what happened: the ring needed no calculus at all, because its symmetry acted on r itself rather than on a cancellation of components.

Once the ring result is in hand, harder geometries become one-dimensional again. Treat the disc as concentric rings: the ring at radius s carries dq = σ 2πs ds and sits at r = √(z² + s²) from the axial point, so dV = k σ 2πs ds / √(z² + s²). The substitution u = z² + s², du = 2s ds turns it into ∫ du/√u, which gives V(z) = 2πkσ (√(z² + R²) − |z|), the same as (σ/2ε₀)(√(z² + R²) − |z|). Put σ = +30.0 nC m⁻², R = 10.0 cm and z = 5.00 cm: σ/2ε₀ = 1694 V m⁻¹, √(z² + R²) = 0.1118 m, and V = 1694 × (0.1118 − 0.0500) = 105 V. The same nesting builds a cylinder out of discs and a sphere out of shells. Each layer contributes a number, so the outer integral stays an ordinary one.

Integrating 1/r along a straight line produces a logarithm, and logarithms misbehave at both ends. For a uniform rod of length L seen end-on from a distance d beyond its near end, V = kλ ∫ dx/(d + x) from x = 0 to L, which is kλ ln((d + L)/d) = kλ ln(1 + L/d). Push d out and ln(1 + L/d) → L/d, so V → kλL/d = kQ/d and the far-field test passes. Push d in and V grows without bound, an artefact of treating real charge as a mathematically thin line. Now let L → ∞ at fixed λ: V diverges at every d, because the idealised line carries infinite charge. The cure is not to abandon the model but to move the reference point. Choose V = 0 at a finite distance r₀ and write V(r) = −2kλ ln(r/r₀) = −(λ/2πε₀) ln(r/r₀). Only potential differences were ever physical, and those stay finite.

Plenty of honest geometries have no elementary antiderivative. Move the field point off the axis of a ring, to a radius b in the plane of the ring, and V = (kQ/2π) ∫ dφ / √(a² + b² − 2ab cos φ) taken over a full turn — a complete elliptic integral. Because the integrand is a number, summing it needs no vector bookkeeping: split the range into N slices, evaluate 1/r at each midpoint, multiply by the slice width and add. For a = 10.0 cm, b = 4.00 cm and Q = +5.00 nC, four slices give 463 V, eight give 469.2 V and sixteen give 469.30 V, already converged. Finish every calculation, closed form or not, with the same three checks: the units must come out in volts, the behaviour close in must be sensible, and at large r the result must fall back onto kQₜₒₜₐₗ/r.